Quiz 3  Acids and Bases  Acids and Bases
General Chemistry 3  Quiz 3  Acids and Bases
What is the conjugate acid of HPO_{4}^{2−}?
The conjugate acid is formed by adding a proton (H^{+}) to the base. For HPO_{4}^{2−}, the conjugate acid is formed by adding one proton, which results in H_{2}PO_{4}^{−}. This species is the conjugate acid of HPO_{4}^{2−}.
What is the pH of a solution made by mixing 200. mL of 0.0657 M NaOH, 140. mL of 0.107 M HCl, and 160. mL of H_{2}O?
Step 1: Calculate the moles of NaOH and HCl

Moles of NaOH: n_{NaOH} = [NaOH] x V_{NaOH} = 0.0657 M x 0.200 L = 0.01314 moles

Moles of HCl: n_{HCl} = [HCl] x V_{HCl} = 0.107 M x 0.140 L = 0.01498 moles
Step 2: Determine the limiting reagent
 NaOH neutralizes HCl in a 1:1 ratio.
 HCl: 0.01498 moles > NaOH: 0.01314 moles
Thus, NaOH is the limiting reagent, and all of it will react with HCl.
Step 3: Calculate the moles of excess HCl after the reaction
After neutralization: Excess HCl moles = 0.01498 moles − 0.01314 moles = 0.00184 moles
Step 4: Determine the concentration of H^{+} ions
The total volume of the solution is: 200 mL + 140 mL + 160 mL = 500 mL = 0.500L
Concentration of H^{+} ions (from excess HCl): [H^{+}] = $\frac{0.00184\mathrm{moles}}{0.500\mathrm{L}}$ = 0.00368 M
Step 5: Calculate the pH
pH= − log [H^{+}] = − log (0.00368) ≈ 2.43
What is the [OH^{–}] of a 0.65 M solution of NaOCl? [Given that K_{a}(HOCl) = 2.8 x 10^{8}]
Step 1: Write the equation for the hydrolysis of NaOCl
OCl^{−} + H_{2}O ⇌ HOCl + OH^{−}
The equilibrium expression for this reaction is based on the base dissociation constant K_{b}.
Step 2: Calculate the K_{b} from K_{a}
We are given the K_{a} of HOCl: K_{a }= 2.8 x 10^{−8}
The relationship between K_{a} and K_{b} for a conjugate acidbase pair is: K_{b} = $\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}}$
where K_{w} is the ionization constant for water (1.0 x 10^{−14} at 25°C). Thus: K_{b} = $\frac{1.0\mathrm{x}{10}^{14}}{2.8\mathrm{x}{10}^{8}}$ = 3.57 x 10^{7}
Step 3: Set up the ICE table and use the K_{b} expression
Let x represent the concentration of OH^{−} at equilibrium.
[OCl^{}]  [HO^{}]  [HOCl]  
Initial  0.65  0  0 
Change  x  x  x 
At equilibrium  0.65  x ≈ 0.65  x  x 
K_{b} = $\frac{\left[\mathrm{HOCl}\right]\left[{\mathrm{HO}}^{}\right]}{\left[{\mathrm{OCl}}^{}\right]}$ = $\frac{{\mathrm{x}}^{2}}{0.65}$
Step 4: Solve for x
x^{2} = (3.57 x 10^{−7}) x 0.65 ≈ 2.32 x 10^{−7} ⇒ x ≈ 4.8 x 10^{−4} M
Thus, the concentration of OH^{−} is approximately 4.8 x 10^{−4} M
What is the percentage ionization of HCOOH molecules in a 0.10 M solution? [K_{a} = 1.8 x 10^{–4}]
Step 1: Write the expression for the ionization of formic acid
The ionization of formic acid can be represented by the equation: HCOOH ⇌ H^{+} + HCOO^{−}
Step 2: Set up the ICE table and expression for K_{a}
Let x represent the concentration of H^{+} and HCOO^{−} ions at equilibrium. The initial concentration of HCOOH is 0.10 M, and it decreases by x upon ionization.
[HCOOH]  [H^{+}]  [HCOO^{}]  
Initial  0.10  0  0 
Change  x  x  x 
At equilibrium  0.10  x  x  x 
K_{a }= $\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{HCOO}}^{}\right]}{\left[\mathrm{HCOOH}\right]}$ = $\frac{{\mathit{x}}^{\mathit{2}}}{\mathit{0}\mathit{.}\mathit{10}\mathit{}\mathit{}\mathit{}\mathit{x}}$ =1.8 x 10^{4}
Since x is expected to be small, we approximate 0.10 − x ≈ 0.10, simplifying the equation:
1.8 x 10^{−4} = $\frac{{x}^{2}}{0.10}$ ⇒ x ≈ 4.24 x 10^{−3 }M
Step 3: Calculate the percentage ionization
The percentage ionization is given by:
Percentage ionization = $\frac{\left[{H}^{+}\right]}{{\left[HCOOH\right]}_{initial}}$ x 100 = $\frac{4.24\mathrm{x}{10}^{3}}{0.10}$ x 100 ≈ 4.2%
The K_{a} of formic acid is 1.8 x 10^{4}. What is the equilibrium constant for the reaction below?
HCO_{2}H + OH^{–} ⇌ HCO_{2}^{–} + H_{2}O
 The reverse reaction is: HCO_{2}^{− }+ H_{2}O ⇌ HCO_{2}H + OH^{−}
The equilibrium constant for this reverse reaction is K_{b}, which is related to the K_{a} of formic acid and the ionization constant of water, K_{w}. 
The relationship between K_{a}, K_{b}, and K_{w} is: Kw = K_{a} x K_{b}
where K_{w} = 1.0 x 10^{14}.
Therefore, K_{b} = $\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}}$ = $\frac{1.0\mathrm{x}{10}^{14}}{1.8\mathrm{x}{10}^{4}}$ = 5.6 x 10^{11} 
The equilibrium constant for the reaction given in the question is the reciprocal of K_{b}, because the reaction is the reverse of the base dissociation reaction. Thus:
K = $\frac{1}{{\mathrm{K}}_{\mathrm{b}}}$ = $\frac{1}{5.6\mathrm{x}{10}^{11}}$ = 1.8 x 10^{10}
What is the pH of a 0.043 M solution of HI?
HI (hydroiodic acid) is a strong acid, which means it fully dissociates in water. Therefore, the concentration of H^{+} ions in solution will be equal to the concentration of the acid itself.
Since HI fully dissociates: [H^{+}] = 0.043 M
pH = − log [H^{+}] = −log (0.043) ≈ 1.37
A 0.10 M solution of glycolic acid is 3.8% ionized. What is the percent ionization of a 0.50 M solution of glycolic acid?
Step 1: Calculate the K_{a} from the given data
The percent ionization of a weak acid is related to the concentration of the acid as follows:
Percent ionization = 3.8% = $\frac{\left[{\mathrm{H}}^{+}\right]}{{\left[\mathrm{HA}\right]}_{0}}$ x 100 with [HA]_{0} = 0.10 M
Let x be the concentration of H^{+} ions. Thus, from the percent ionization formula:
[H^{+}] = x = $\frac{3.8}{100}$ x 0.10 = 3.8 x 10^{3} M
Now, let's calculate K_{a} for glycolic acid:
K_{a} = $\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{}\right]}{\left[\mathrm{HA}\right]}$ = $\frac{{\mathrm{x}}^{2}}{0.10\mathrm{x}}$ = 1.444 x 10^{4}
Step 2: Calculate [H^{+}] in the 0.50M solution
Now, for the 0.50M solution, we can use the same K_{a} value calculated from the first step:
K_{a} = $\frac{{\left[{\mathrm{H}}^{+}\right]}^{2}}{0.50}$ ⇒ [H^{+}] = $\sqrt{0.50{\mathrm{K}}_{\mathrm{a}}}$ = $\sqrt{7.22\times {10}^{5}}$ = 8.5 x 10^{3} M
Step 3: Calculate the % ionization in the 0.50M solution
The percent ionization in the 0.50 M solution is:
Percent Ionization = $\frac{8.5\times {10}^{3}}{0.50}$ x 100 = 1.7%
A solution of ammonia, NH_{3}, has pH = 11.50. What is the ammonia concentration? (The pK_{a} of NH_{4}^{+} is 9.24)
Step 1: Calculate the [OH^{−}]
Given that the pH is 11.50, we can calculate the pOH first:
pOH = 14 − pH = 14 − 11.50 = 2.50
Next, calculate the concentration of hydroxide ions: [OH^{−}] = 10^{−pOH }= 10^{−2.50 }≈ 3.16 x 10^{−3} M
Step 2: Use the relationship between K_{b} and K_{a}
We are given the pK_{a} of NH_{4}^{+}, which is 9.24. Using the relationship between K_{a} and K_{b}:
K_{b} = $\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}}$ = $\frac{1.0\mathrm{x}{10}^{14}}{{10}^{9.24}}$ = 5.75 x 10^{5}
Step 3: Set up the equilibrium expression for ammonia
The reaction for ammonia in water is: NH_{3} + H_{2}O ⇌ NH_{4}^{+} + OH^{−}
The expression for the base dissociation constant K_{b} is: K_{b} = $\frac{\left[{{\mathrm{NH}}_{4}}^{+}\right]\left[{\mathrm{OH}}^{}\right]}{\left[{\mathrm{NH}}_{3}\right]}$
Assuming that the concentration of [NH_{4}^{+}] is approximately equal to [OH^{−}] at equilibrium (since it comes from the same ionization reaction), we can substitute:
K_{b} = $\frac{{\left[{\mathrm{OH}}^{}\right]}^{2}}{\left[{\mathrm{NH}}_{3}\right]}$ ⇒ [NH_{3}] = $\frac{{\left[{\mathrm{OH}}^{}\right]}^{2}}{{\mathrm{K}}_{\mathrm{b}}}$ = $\frac{{(3.16\mathrm{x}{10}^{3})}^{2}}{5.75\mathrm{x}{10}^{5}}$ ≈ 0.58 M